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AVprozaik [17]
3 years ago
5

If the tire reaches a temperature of 48 ∘c, what fraction of the original air must be removed if the original pressure of 250 kp

a is to be maintained?
Physics
1 answer:
andrey2020 [161]3 years ago
7 0
Assuming air as ideal gas and amount of air in no of moles is known then by gas law,

PV= nRT

Pressure is constant

P* (change in volume) = nR* (change in temperature)
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v = 90.1 m/s
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Who made advances to our understanding of the conscious and unconscious mind?
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What is one standard kilogramun si system<br><br><br><br><br>​
Phoenix [80]

Answer:

The kilogram (kg) is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015 ×10−34 when expressed in the unit J s, which is equal to kg m2 s−1, where the meter and the second are defined in terms of c and ∆νCs.

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Explain the difference between the synodic period of a planet and the sidereal period. why is the synodic period often much less
Ilia_Sergeevich [38]
-Synodic period is the period of celestial bodies observed on the moving planet(mostly earth)
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The engine in an imaginary sports car can provide constant power to the wheels over a range of speeds from 0 to 70 miles per hou
Mama L [17]

A) Time needed: 6.24 s

B) Time needed: 2.86 s

Explanation:

A)

In this part, we are told that the power if the engine is constant. The power of the engine is given by

P=\frac{W}{t}

where

W is the work done

t is the time

This means that the power of the engine is proportional to the work done, and therefore, to the kinetic energy of the car:

P=\frac{\frac{1}{2}mv^2}{t}=const.

where m is the mass of the car and v its velocity.

SInce power is constant, we can write:

\frac{\frac{1}{2}mv_1^2}{t_1^2}=\frac{\frac{1}{2}mv_2^2}{t_2}

where:

t_1=1.40 s is the time the car needs to accelerates to v_1=28.0 mph

t_2 is the time the car needs to accelerate to v_2=57.0 mph

Therefore, solving for t_2,

t_2 = \frac{v^2}{u^2}t_1=\frac{57^2}{28^2}(1.40)=6.24 s

B)

First of all, we have to calculate the acceleration of the car. We can do it using the following equation:

a=\frac{v-u}{t}

where:

u = 0 is the initial velocity

v=28.0 mph \cdot \frac{1609 m/mi}{3600 s/h}=12.5 m/s is the final velocity

t = 1.40 s is the time elapsed

Substituting, we find the acceleration:

a=\frac{12.5-0}{1.40}=8.9 m/s^2

In this part, we are told that the force exerted by the engine is constant: according to Newton's second law, acceleration is proportional to the force,

F=ma

This means that the acceleration is also constant.

Now we want to find how long the car takes to accelerate to a final velocity of

v=57.0 mph \cdot \frac{1609}{3600}=25.5 m/s

From an initial velocity of

u = 0

Using again the same suvat equation, and using the acceleration we found previously, we find:

t=\frac{v-u}{a}=\frac{25.5-0}{8.9}=2.87 s

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

About power:

brainly.com/question/7956557

#LearnwithBrainly

6 0
3 years ago
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